I'll factor it...
125 + 8y^3 = (5^3) + (2y)^3
This is the sum of two cubes
factors to: (5 + 2y)(25 - 10y + 4y^2)
Check by multiplying
QED
Better recall a^3 + b^3 = (a+b)(a^2--ab+b^2)
Here a = 5 and b = 2y
So 125 + 8y^3 = (5+2y)(25--10y+4y^2)
125 + 8y^3 = ( 5 + 2*y )*( 25 - 10*y + 4*y^2 )
a^3 + b^3 ≡ (a + b)(a^2 - ab + b^2)
125 + 8y^3
= 5^3 + (2y)^3
= (5 + 2y)(25 - 10y + 4y^2)
125 + 8y^3 is a sum of cubes. (a^3 + b^3) = (a+b)(a^2 - ab + b^2)
= (5+2y)(25-10y+4y^2)
125+8y^3
= 5^3 + (3y)^3
= (5+2y)(25+4y^3 - 10y) [reason: (a+b)^3=(a+b) (a^2 + b^2 - ab)]
Copyright © 2024 EBIN.TIPS - All rights reserved.
Answers & Comments
Verified answer
I'll factor it...
125 + 8y^3 = (5^3) + (2y)^3
This is the sum of two cubes
factors to: (5 + 2y)(25 - 10y + 4y^2)
Check by multiplying
QED
Better recall a^3 + b^3 = (a+b)(a^2--ab+b^2)
Here a = 5 and b = 2y
So 125 + 8y^3 = (5+2y)(25--10y+4y^2)
125 + 8y^3 = ( 5 + 2*y )*( 25 - 10*y + 4*y^2 )
a^3 + b^3 ≡ (a + b)(a^2 - ab + b^2)
125 + 8y^3
= 5^3 + (2y)^3
= (5 + 2y)(25 - 10y + 4y^2)
125 + 8y^3 is a sum of cubes. (a^3 + b^3) = (a+b)(a^2 - ab + b^2)
= 5^3 + (2y)^3
= (5+2y)(25-10y+4y^2)
125+8y^3
= 5^3 + (3y)^3
= (5+2y)(25+4y^3 - 10y) [reason: (a+b)^3=(a+b) (a^2 + b^2 - ab)]